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1 ) ____ P 4 + ____ O 2  ____ P 2 O 3 2 ) ____ RbNO 3 + ____ BeF 2  ____ Be(NO 3 ) 2 + ____ RbF 3 ) ____ AgNO 3 + ____ Cu  ____ Cu(NO 3 ) 2 + ____ Ag 4 ) ____ CF 4 + ____ Br 2  ____ CBr 4 + ____ F 2 5 ) ____ HCN + ____ CuSO 4  ____ H 2 SO 4 + ____ Cu(CN) 2. DO NOW.

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DO NOW

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  1. 1) ____ P4 + ____ O2 ____ P2O3 2) ____ RbNO3 + ____ BeF2 ____ Be(NO3)2 + ____ RbF 3) ____ AgNO3 + ____ Cu  ____ Cu(NO3)2 + ____ Ag 4) ____ CF4 + ____ Br2 ____ CBr4 + ____ F2 5) ____ HCN + ____ CuSO4 ____ H2SO4 + ____ Cu(CN)2 DO NOW

  2. After today you will be able to… • Calculate the number of moles of a substance that can be produced from their coefficients in the balanced chemical equation • Review of the basics of how to balance a chemical equation

  3. The wordstoichiometycomes from the Greek words: stoicheion(meaning "element") metron(meaning "measure") Stoichiometry: The study of quantities as it relates to chemical reactions.

  4. Balanced chemical equations can be interpreted many ways… • 1N2(g) + 3H2(g) 2NH3(g) H • molecules: N N H H H H H H H H H H N N +  H

  5. Balanced chemical equations can be interpreted many ways… • 1N2(g) + 3H2(g) 2NH3(g) • mass: +  • 34.08g • 28.02g • 6.06g • 34.08g • = • 34.08g • Mass of reactants and products are always equal!

  6. Balanced chemical equations can be interpreted many ways… • 1N2(g) + 3H2(g) 2NH3(g) • moles: +  • 2 mol NH3 • 1 mol N2 • 3 mol H2 • This is the relationship we will focus on in this unit!

  7. Mole-Mole Calculations • The coefficients in the balanced equation represent the smallest whole number mole ratio between reactants and products.

  8. Mole-Mole Calculations • Examples of mole ratios for the reaction: • 1N2(g) + 3H2(g) 2NH3(g) 1 mol N2 2 mol NH3 3 mol H2 1 mol N2 2 mol NH3 2 mol H2

  9. Mole-Mole Calculations • Example: If 3.86 moles of potassium reacts completely with excess water, how many moles of hydrogen would be produced? hydrogen hydroxide water potassium +  potassium hydroxide + hydrogen __K 2 __ H(OH) 2 __ K(OH) + __ H2 1 2  + K: 3.86 mol K U: ? mol H2 3.86 mol K 1 1 mol H2 2 mol K 1.93 mol H2 x =

  10. Mole-Mole Calculations • Example: How many moles of aluminum will react with 0.512 moles of hydrochloric acid? hydrogen chloride aluminum + hydrogen chloride aluminum + hydrochloric acid  __Al 2 __HCl 6 __ AlCl3 + __ H2 3 2  + K: 0.512 mol HCl U: ? mol Al 2 mol Al 6 mol HCl 0.512 mol HCl 1 0.171 mol Al x =

  11. Complete Worksheet! Questions?

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